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+//chapter11
+//example11.13
+//page211
+
+Ic=1000 // micro ampere
+// when emitter circuit is open, leakage current = Icbo so
+Icbo=0.2 // micro ampere
+
+// when base is open, leakage current = Iceo so
+Iceo=20 // micro ampere
+
+//since Iceo=Icbo/(1-alpha) we get
+alpha=1-(Icbo/Iceo)
+
+// since Ic=alpha*Ie+Icbo we get
+Ie=(Ic-Icbo)/alpha
+Ib=Ie-Ic
+
+printf("alpha = %.3f \n",alpha)
+printf("emitter current = %.3f micro ampere \n",Ie)
+printf("base current = %.3f micro ampere \n",Ib)